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3 years ago

Help me on number problem 1 please

Mathematics

1 answer:

Mandarinka [93]3 years ago

6 0

Answer: y = 3x-1

Step-by-step explanation: I was never the best at doing this, but here it goes...

first start with the formula Y2-Y1/X2-X1 for this problem, lets say your Y2 is -1, your Y1 is 5, your X2 is 0, and your X1 is 2, making your equation -1-5/0-2 which equals -6/-2, which can be qritten as 3. this is your slope.

now take the equation Y-Y1 = M (X-X1) m is your slope, which is 3 in this case as stated above. your X1 is2, your Y1 is 5. plug this in and you get

y-5 = 3 (x-2) which you put into slope intercept form, which is y = mx + b

your answer should be y = 3x-1

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3 years ago

8 + 3 + 2a = -1 + 3a + 4

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Is the function even odd or neither?

Mandarinka [93]

The answer is the first option: Even.

The explanation for this exercise is shown below:

1. By definition, if $f(x)=f(-x)$ the fucntion is even.

2. When the graph is symmetric with respect to the y-axis, it is an even function.

3. As you you can see in the graph attached in the problem, the graph is symmetric about the y-axis. Therefore, you can conclude it is an even function.

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3 years ago

A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horiz

Alex73 [517]

Answer:

Step-by-step explanation:

In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

$-15=(0)t+\frac{1}{2}(-32)t^2$ which simplifies down to

$-15=-16t^2$ so

$t=\sqrt{\frac{-15}{-16} }$ so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = $(80)(.968) +(0)(.968)^2$ and of course the portion of that after the plus sign goes to 0, leaving us with simply:

Δx = (80)(.968)

Δx = 77.46 feet

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3 years ago